Toán Lớp 8: (25x-15)/(1-25x^(2))+(1)/(x-5x^(2))
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Toán Lớp 8: (25x-15)/(1-25x^(2))+(1)/(x-5x^(2))
Comments ( 2 )
Giải đáp:
(25x – 15)/(1 – 25x^2) + 1/(x – 5x^2) (x \ne 0; ± 1/5)
= (25x – 15)/[(1 – 5x).(1 + 5x)] + 1/[x.(1 – 5x)]
= [x.(25x – 15)]/[x.(1 – 5x).(1 + 5x)] + [1 + 5x]/[x.(1 – 5x).(1 + 5x)]
= [25x^2 – 15x]/[x.(1 – 5x).(1 + 5x)] + [1 + 5x]/[x.(1 – 5x).(1 + 5x)]
= [25x^2 – 15x + 1 + 5x]/[x.(1 – 5x).(1 + 5x)]
= [1 – 10x + 25x^2]/[x.(1 – 5x).(1 + 5x)]
= [(1 – 5x)^2]/[x.(1 – 5x).(1 + 5x)]
= [1 – 5x]/[x.(1 + 5x)]
$#dariana$
Giải đáp:
(1 – 5x)/(x(1 + 5x))
Lời giải và giải thích chi tiết:
(25x – 15)/(1 – 25x^2) + 1/(x – 5x^2) (ĐK: x \ne 0, x \ne +-1/5)
= (25x – 15)/(1 – (5x)^2) + 1/(x(1 – 5x))
= (25x – 15)/((1 – 5x)(1 + 5x)) + 1/(x(1 – 5x))
= (x(25x – 15))/(x(1 – 5x)(1 + 5x)) + (1(1 + 5x))/(x(1 – 5x)(1 + 5x))
= (x(25x – 15) + 1 + 5x)/(x(1 – 5x)(1 + 5x))
= (25x^2 – 15x + 1 + 5x)/(x(1 – 5x)(1 + 5x))
= (1 – 10x + 25x^2)/(x(1 – 5x)(1 + 5x))
= (1 – 5x)^2/(x(1 – 5x)(1 + 5x))
= (1 – 5x)/(x(1 + 5x))